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TLDR: Is $ |Isom(Z_p)| = 2^{\mathfrak c} $?

For simplicity let $ Isom(\mathbb Z_p) = \{ \mathbb Z_p \xrightarrow{f} \mathbb Z_p \mid f \text{ is an isometry} \} $.

How to compute $ \left| Isom(\mathbb Z_p) \right| $? It's clear that there are many affine isometries of the form: $$ f(x) = a x + b, \quad a \in \mathbb Z^\times_p, b \in \mathbb Z_p $$

This suggests that the cardinality of $ Isom(\mathbb Z_p​) $ satisfies $$ \mathfrak{c} \leqslant |Isom(\mathbb Z_p)| \leqslant 2^{\mathfrak c}, $$

where $ \mathfrak{c} $ is the cardinality of the continuum.

On the other hand, I can build many "digit-manipulating" isometries, but I can't come up with neither really large family that will prove $ |Isom(\mathbb Z_p)| = 2^{\mathfrak c} $ nor strict bound $ |Isom(\mathbb Z_p)| < 2^{\mathfrak c} $.

Izaak van Dongen
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tensorix
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  • Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – Sigma Algebra Aug 21 '24 at 16:46
  • @SigmaAlgebra added TLDR section. Is it better now? – tensorix Aug 21 '24 at 16:58
  • Yes ok, but you need to add your work to the post so that users can understand what you are doing and where you have doubts. Questions with only the problem are not given much consideration. – Sigma Algebra Aug 21 '24 at 17:07
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    Hello. Nice question! I think your question is perfectly clear, and I would say you have shown enough context and work. Here is a hint: take a few steps back and first ask yourself how many continuous functions there even are on $\Bbb Z_p$. This problem is actually very similar to the same problem for $\Bbb R$ itself, so maybe you'd find it easier to think about that question. If you don't mind just spoiling it for yourself, there are answers here. – Izaak van Dongen Aug 23 '24 at 11:10
  • @JyrkiLahtonen nice catch! Completely forgot that basic argument! This is actually an answer – tensorix Aug 23 '24 at 14:46

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You can argue analogously to the answers in the similar thread about the cardinality of the set of real continuous functions. Every $p$-adic integer $m$ is a limit of a sequence $(m_i)_{i\in\Bbb{N}}$ of integers $\in\Bbb{Z}\subset\Bbb{Z}_p$. An isometry $f$ necessarily maps a Cauchy sequence to another, and hence $f(m)=\lim_{i\to\infty}f(m_i)$. This means that $f$ is fully determined, if we know its restricion $f\vert_{\Bbb{Z}}$. The set of sequences $\Bbb{Z_p}^{\Bbb{Z}}$ has continuum cardinality, so the set of isometries cannot have a higher cardinality. The OP already showed the converse inequality.

Jyrki Lahtonen
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